My question is the following:
Let M be a simply connected Riemannian manifold whose sectional curvatures are all nonpositive and let G be a group. Suppose that G acts in M properly discontinuous and cocompactly by isometries. Is G a CAT(0) group?
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1 Answer
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The answer is yes, since $M$ is a $CAT(0)$ space, and the group is quasi-isometric to it. See (for example) Jim Cannon's article in Bedford Keane Series:
The theory of negatively curved spaces and groups J Cannon - t. Bedford, M. Keane, and C. Series. Oxford University …, 1991
answered Dec 10, 2011 at 16:16
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2$\begingroup$ By definition, a group is $CAT(0)$ if it acts isometrically, properly and co-compactly on a $CAT(0)$ space... So it is enough to say that $M$ is $CAT(0)$. See aimath.org/pggt/… $\endgroup$ Commented Dec 10, 2011 at 18:33
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$\begingroup$ Igor, you should say "yes by definition", but not "yes, since". $\endgroup$– ε-δCommented Dec 10, 2011 at 21:22
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$\begingroup$ @ε-δ: I believe that is the essence of Alain's commentary... $\endgroup$ Commented Dec 11, 2011 at 10:08
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$\begingroup$ To be fair, it is not a trivial fact that a simply connected Riemannian manifold whose sectional curvatures are all nonpositive is CAT(0). $\endgroup$– AnonCommented Dec 15, 2011 at 10:01
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3$\begingroup$ @Anon: I believe that this is why the T is in CAT(0)... $\endgroup$ Commented Dec 15, 2011 at 10:37